In mathematics, telescoping refers to a property of certain algebraic expressions or iterated binary operations in which successive terms cancel each other after expansion. As a result, the overall expression simplifies significantly, since most intermediate terms eliminate one another, leaving only a small number of boundary terms from the original structure. Most commonly, expressions of the form
f ( n ) = g ( n + 1 ) − g ( n ) {\displaystyle f(n)=g(n+1)-g(n)}
or
f ( n ) = g ( n + 1 ) g ( n ) , g ( n ) ≠ 0 {\displaystyle f(n)={\frac {g(n+1)}{g(n)}},\quad g(n)\neq 0}
are considered, where g {\displaystyle g} is a function defined on the set of natural numbers.
Definition Let A ⊆ Z {\displaystyle A\subseteq \mathbb {Z} } and let f , g : A → R {\displaystyle f,g:A\to \mathbb {R} } . We say that a function f {\displaystyle f} has the telescoping property with respect to g {\displaystyle g} if one of the following holds:
Additive case
f ( n ) = g ( n + 1 ) − g ( n ) {\displaystyle f(n)=g(n+1)-g(n)} . Then the sum
∑ k = m n f ( k ) {\displaystyle \sum _{k=m}^{n}f(k)}
is telescoping.
Multiplicative case If g ( n ) ≠ 0 {\displaystyle g(n)\neq 0} , then
f ( n ) = g ( n + 1 ) g ( n ) {\displaystyle f(n)={\frac {g(n+1)}{g(n)}}} . Then the product
∏ k = m n f ( k ) {\displaystyle \prod _{k=m}^{n}f(k)}
reduces telescopically.
Relation to finite difference operator Let Δ {\displaystyle \Delta } be the finite difference operator:
Δ g ( n ) = g ( n + 1 ) − g ( n ) {\displaystyle \Delta g(n)=g(n+1)-g(n)} . Then in the additive case:
f ( n ) = Δ g ( n ) {\displaystyle f(n)=\Delta g(n)} . This is analogous to the Newton–Leibniz formula:
∑ k = m n Δ g ( k ) = g ( n + 1 ) − g ( m ) {\displaystyle \sum _{k=m}^{n}\Delta g(k)=g(n+1)-g(m)} . Similarly for products:
∏ k = m n g ( k + 1 ) g ( k ) = g ( n + 1 ) g ( m ) , g ( k ) ≠ 0 {\displaystyle \prod _{k=m}^{n}{\frac {g(k+1)}{g(k)}}={\frac {g(n+1)}{g(m)}},\quad g(k)\neq 0} .
Examples
Telescoping sum
f ( n ) = 1 n ( n + 1 ) {\displaystyle f(n)={\frac {1}{n(n+1)}}}
Decomposition:
f ( n ) = 1 n − 1 n + 1 {\displaystyle f(n)={\frac {1}{n}}-{\frac {1}{n+1}}}
so for g ( n ) = − 1 n {\displaystyle g(n)=-{\frac {1}{n}}} we have
f ( n ) = g ( n + 1 ) − g ( n ) {\displaystyle f(n)=g(n+1)-g(n)} . Thus:
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